A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is
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a
4
b
6
c
8
d
12
answer is C.
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Detailed Solution
n∝1l⇒Δnn=−ΔllIf length is decreased by 2% then frequency increases by 2% i.e., n2−n1n1=2100⇒n2−n1=2100×n1=2100×392=7.8≈8.